Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Re: constructing an example
Posted:
Jul 27, 2011 4:57 PM


In article <folt27pu1l3bv9efboq189203ucr37pl79@4ax.com>, David C. Ullrich <ullrich@math.okstate.edu> wrote: >On Mon, 25 Jul 2011 11:22:35 0700 (PDT), Dan <dan.ms.chaos@gmail.com> >wrote: > >>Let g : [0,1) > R be >>g(x) = 1/2 for x <= 1/2 >>g(x) = 1/2 for x> 1/2 >> >>u(x) = (2*x ) mod 1 >> >>Can h(x) be constructed so that h(u(x))  h(x) is equivalent to g(x) >>in LP2 ? >>(integral from 0 to 1 over x of ((h(u(x))  h(x) )  g(x) )^2 = >>0 ) > >One can save typing by replacing your g by 2g. > >So the question is does there exist h such that h(2x)  h(x) = 1 >almost everywhere for 0 < x < 1/2 and h(2x  1)  h(x) = 1 >almost everywhere for 1/2 < x < 1. > >It seems to me that there is no such h in L^2 (I gather your >"LP2" is the space commonly known as L^2?). If you say >c_n is the nth Fourier coefficient of h then it seems to me >that (dropping irrelevant constants) you'd need >c_n = 1/n for n odd and c_{2n} = c_n. > >Except that c_0 is arbitrary, there's exactly one sequence >that satisfies those two conditions. It doesn't tend to 0, >so h cannot be L^2 or even L^1. On the other hand, >that sequence is bounded, so there _is_ a 2piperiodic >diastribution that does what you ask, at least formally. >Whether that distribution can be given as a locally >integrable function, or whether h can be taken to be >measurable and satisfy your condition pointwise, is >not immediately clear to me.
If we don't drop the irrelevant constants, we get that the fourier series for h(x) would formally be
+oo 2  h(x) =  > c sin(2?kx) ?  k k=1
where c_k is the reciprocal of the largest odd divisor of k.
Since the fourier series of g(x) is
+oo 2  1 g(x) =   >  sin(2?(2k+1)x) ?  2k+1 k=0
We get that, formally,
+oo  k h(x) =  > g(2 x)  k=0
Here, at least formally, h(2x)  h(x) = g(x).
If we try to compute h(x) for a given x which doesn't have a finite binary expansion, we get that h(x) is the sum over the binary digits of x, 1/2 for 0 and 1/2 for 1. My bet is against there being a locally integrable function for h.
Consider h as the limit as n > oo of
n  k h (x) =  > g(2 x) n  k=0
Then, it is pretty straightforward to show that h is bounded as a distribution on Lip(?) for ? > 0, with a constant of 1/(2^?  1), which is about 1/(log(2) ?) when ? is near 0.
Rob Johnson <rob@trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font



